The Stacks project

Proof. If $x \in |X|$ is closed, then we can represent $x$ by a closed immersion $\mathop{\mathrm{Spec}}(k) \to X$, see Lemma 68.14.6. Hence $x$ is certainly a finite type point.

Conversely, let $x \in |X|$ be a finite type point. We know that $x$ can be represented by a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k) \to X$ where $k$ is a field (Definition 68.6.1). On the other hand, by definition, there exists a morphism $\mathop{\mathrm{Spec}}(k') \to X$ which is locally of finite type and represents $x$ (Morphisms, Definition 29.16.3). We obtain a factorization $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k) \to X$. Let $U \to X$ be any étale morphism with $U$ affine and consider the morphisms

The quasi-compact scheme $\mathop{\mathrm{Spec}}(k) \times _ X U$ is étale over $\mathop{\mathrm{Spec}}(k)$ hence is a finite disjoint union of spectra of fields (Remark 68.4.1). Moreover, the first morphism is surjective and locally of finite type (Morphisms, Lemma 29.15.8) hence surjective on finite type points (Morphisms, Lemma 29.16.6) and the composition (which is locally of finite type) sends finite type points to closed points as $U$ is Jacobson (Morphisms, Lemma 29.16.8). Thus the image of $\mathop{\mathrm{Spec}}(k) \times _ X U \to U$ is a finite set of closed points hence closed. Since this is true for every affine $U$ and étale morphism $U \to X$, we conclude that $x \in |X|$ is closed. $\square$